# I load the needed libraries
library(dplyr)
library(scales)
library(ggplot2)
library(rjags)
library(coda)

Attaching package: ‘dplyr’


The following objects are masked from ‘package:stats’:

    filter, lag


The following objects are masked from ‘package:base’:

    intersect, setdiff, setequal, union


The following objects are masked from ‘package:stats’:

    filter, lag


The following objects are masked from ‘package:base’:

    intersect, setdiff, setequal, union


Loading required package: coda

Linked to JAGS 4.3.2

Loaded modules: basemod,bugs

Exercise 1

Given the following un-normalized posterior distribution:

$$ g \left(\theta|x\right) \propto \frac{1}{2} e^{-\frac{\left(\theta + 3\right)^2}{2}} + \frac{1}{2} e^{-\frac{\left(\theta - 3\right)^2}{2}} $$

• Draw a Markov Chain from the posterior distribution using a Metropolis-Hastings algorithm using a $Norm (0, 1)$ as random-walk candidate density and finally plot the sampled distribution
• Analyze the chain with the CODA package and plot the chain autocorrelation
• Try to use different burn-in cycles and thinning and plot the corresponding posterior distribution and the chain autocorrelation function. What are the best parameters ?

$\bold{a)}$ I use the same functions created in Exercise 2 R Lab 5 to create the MCMC in a "homemade" way:

# Function to sampled from:
posterior_g = function (theta) {
    return (0.5*(exp(-((theta + 3)**2)/2) + exp(-((theta - 3)**2)/2)))
}

# This function is ment to return the sequence of the samples for a determined function
random_steps = function (func_wanted, theta_init, n_samples, func_change, sigma, print_accept=FALSE) {
    # Initilalizing the parameters
    current_theta = theta_init
    current_function = func_wanted(theta_init)
    samples = matrix(data = NA, nrow = n_samples, ncol = 2)

    # For statistical purposes
    accepted = 0

    # Evolution loop
    for (n in 1:n_samples) {
        # Take a guessed new theta and evaluate its probability
        guessed_theta = func_change(current_theta, sigma)
        guessed_function = func_wanted(guessed_theta)

        # Acceptance conditions
        if ((guessed_function > current_function) || ((guessed_function/current_function) > runif(1))) {
            current_theta = guessed_theta
            current_function = guessed_function
            accepted = accepted + 1
        }

        # Saving the generated samples
        samples[n, 1] = current_function
        samples[n, 2] = current_theta
    }

    if(print_accept) {
        cat("Acceptance rate = ", round(accepted/n_samples, 5), '\n')
    }

    return(samples)
}

# I have to use a gaussian function to generate the guessed new parameters in the MCMC
generation_function = function (x0, std) {
    return(rnorm(n = 1, mean = x0, sd = std))
}

# The initial parameters are:
init = 1; std = 1; burn_in = 1000
N = as.integer(1e5) + burn_in

mcmc_g = random_steps(func_wanted = posterior_g, theta_init = init, n_samples = N, func_change = generation_function, sigma = std, print_accept=TRUE)
mcmc_g = mcmc_g[,2][burn_in:N] # Selecting the sequence after the burn-in

Acceptance rate =  0.71099 

# Printing then the evolution of the chain and the resulting distribution
par(mfrow=c(1,2), oma = c(0, 0, 0, 0))
options(repr.plot.width=20, repr.plot.height=6)

steps = 100

plot_g = mcmc_g[seq(0, length(mcmc_g), steps)] # I take a burn-in period
plot(1:length(plot_g)*steps, plot_g, type = 'o', lwd = 1, col = 'black', xlab = 'Iteration', ylab = 'Theta evolution',
        main = paste0('Initial value = ', init, '     Standard deviation = ', std))
hist(mcmc_g, breaks = 30, xlab = 'Lambda', ylab = 'Counts', main = 'Histogram of the posterior distribution')

$\bold{b)}$ Now I analize the chain using the CODA package:

# I ensure that this is now a MCMC
g_chain = as.mcmc(mcmc_g)

# Then I check the autocorrelation using CODA
lags = seq(0,1000,10)
auto_g = autocorr(g_chain, lags=lags)

# And finally I plot the autocorrelation evolution:
plot(lags, auto_g, ylim=c(0,1), pch=7, col="navy", xlab="lag", ylab="ACF", cex=1.5, main = paste("Autocorrelation behaviour with sigma =", std))
text(900,0.9, sprintf (" effective size: %.2f", effectiveSize(g_chain)))

$\bold{c)}$ I loop over the dimension of the sigma in order to obtain the best parameters, using the functions defined before

# Group of sigmas to loop on:
sigmas = c(0.1, 0.5, 1, 2, 3, 5, 10)

# Other parameters useful to be defined again:
init = 1; std = 1; burn_in = 1000
N = as.integer(1e5) + burn_in
lags = seq(0,1000,10)


for (std in sigmas) {
    # I recicle the functions used before:
    mcmc_g = random_steps(func_wanted = posterior_g, theta_init = init, n_samples = N, func_change = generation_function, sigma = std, print_accept=FALSE)
    mcmc_g = mcmc_g[,2][burn_in:N] # Selecting the sequence after the burn-in
    g_chain = as.mcmc(mcmc_g)
    auto_g = autocorr(g_chain, lags=lags)

    # And then I plot the results
    par(mfrow=c(1,3), oma = c(0, 0, 5, 0))
    options(repr.plot.width=20, repr.plot.height=6)
    steps = 100

    plot_g = mcmc_g[seq(0, length(mcmc_g), steps)] # I take a burn-in period
    plot(1:length(plot_g)*steps, plot_g, type = 'o', lwd = 1, col = 'black', xlab = 'Iteration', ylab = 'Theta evolution', main = 'Evolution of the chain')
    hist(mcmc_g, breaks = 30, xlab = 'Lambda', ylab = 'Counts', main = 'Histogram of the posterior distribution')
    plot(lags, auto_g, ylim=c(0,1), pch=7, col="navy", xlab="lag", ylab="ACF", cex=1.5, main = paste("Autocorrelation behaviour"))

    mtext(paste("Evolution of the chain with sigma =", std) , outer = TRUE, cex = 2, font = 1)
}

From these plots we see that one obtains good results with a sigma greater than one: this means that the best parameter is $\sigma = 1$

Exercise 2

the European Medicines Agency (EMA) has authorized a list of COVID-19 vaccines, after having performed a scientific evaluation of the veccines efficacy. The following vaccines are currently authorized for use in the European Union:

• Comirnaty (BioNTech and Pfizer)
• VCOVID-19 Vaccine Valneva
• Nuvaxovid (Novavax)
• Pikevax (Moderna)
• Vaxzeviria (AstraZeneca)
• Jcovden (Janssen)
• VidPrevtyn Beta (Sanofi Pasteur)
• Bimervax, previously COVID-19 Vacxcine HIPRA (HIPRA Human Health S.L.U.)

Analyze the initial test data reported on the EMA Web site for the following early Vaccines:

• Janssen [https://www.ema.europa.eu/en/documents/overview/covid-19-vaccine-janssen-epar-medicine-overview_en.pdf]
• Moderna [https://www.ema.europa.eu/en/documents/overview/spikevax-previously-covid-19-vaccine-moderna-epar-medicine-overview_en.pdf]
• AstraZeneca [https://www.ema.europa.eu/en/documents/overview/covid-19-vaccine-astrazeneca-epar-medicine-overview_en.pdf]
• Jcovden [https://www.ema.europa.eu/en/documents/overview/jcovden-previously-covid-19-vaccine-janssen-epar-medicine-overview_en.pdf]

Then create a Markow Chain Monte Carlo (using JAGS or stan) and evaluate the efficacy of each Vaccine. Infere the 95% credibility interval.

$\bold{a)}$ After loading the data, I can create the MCMC using JAGS (and then connecting a .bug file) for the JANSSEN vaccine:

# I save the relevant data for the 
vaccine = 19630
placebo = 19691
vaccine_pos = 116
placebo_pos = 348

total = c(rep("V", vaccine), rep("P", placebo)) # V = vaccined
positives = c(rep("P", vaccine_pos), rep("N", vaccine - vaccine_pos), rep("P", placebo_pos), rep("N", placebo - placebo_pos)) # 1 = positive to COVID

# Then I can create the samples list containing these data
data = list(
    total = as.integer(factor(total)),
    length_total = length(unique(total)),
    positives = ifelse(positives == "N", 0, 1),
    length_positives = length(positives)
)

# And then I can use the .bug file to create the posterior
jags_model = jags.model('ex02_les06.bug', data)
update(jags_model, 1000)

# Creation of the rjags chain
janssen_chain = coda.samples(jags_model, c("distribution"), n.iter=10000)
print(summary(janssen_chain))

# And in the end I create a plot for clarity of the results
options(repr.plot.width=20, repr.plot.height=8)
plot(janssen_chain, col="navy")

Compiling model graph
   Resolving undeclared variables
   Allocating nodes
Graph information:
   Observed stochastic nodes: 39321
   Unobserved stochastic nodes: 2
   Total graph size: 78648

Initializing model


Iterations = 1001:11000
Thinning interval = 1 
Number of chains = 1 
Sample size per chain = 10000 

1. Empirical mean and standard deviation for each variable,
   plus standard error of the mean:

                    Mean        SD  Naive SE Time-series SE
distribution[1] 0.017712 0.0009470 9.470e-06      9.470e-06
distribution[2] 0.006042 0.0005536 5.536e-06      5.536e-06

2. Quantiles for each variable:

                   2.5%      25%     50%      75%    97.5%
distribution[1] 0.01591 0.017049 0.01770 0.018332 0.019605
distribution[2] 0.00499 0.005658 0.00603 0.006414 0.007165

# For the 95% credibility interval and the efficacy I load both the chain and the result file
chain_df = as.data.frame( as.mcmc(janssen_chain) )
results = summary(janssen_chain)

# 95% CI:
cat("The Credibility interval is [", round(results$quantiles[1,1], 4), ",", round(results$quantiles[1,5], 4),"] for the Vaccined distribution\n")
cat("The Credibility interval is [", round(results$quantiles[2,1], 5), ",", round(results$quantiles[2,5], 5),"] for the Placeboed distribution\n")

# For plotting reasons I retrieve mean, variance and standard deviation of the chain
chain_means = sapply(chain_df, mean)
chain_vars = sapply(chain_df, var)
chain_stds = sqrt(chain_vars)

# Efficacy
cat("The efficacy of the test is ", round((1 - chain_means[2]/chain_means[1])*100, 2), "%")

The Credibility interval is [ 0.0159 , 0.0196 ] for the Vaccined distribution
The Credibility interval is [ 0.00499 , 0.00716 ] for the Placeboed distribution
The efficacy of the test is  65.89 %The Credibility interval is [ 0.00499 , 0.00716 ] for the Placeboed distribution
The efficacy of the test is  65.89 %

Here I see that I obtain the same results given by the EMA autorities

$\bold{a)}$ Now I perform the same analysis for the MODERNA vaccine, based on the extracted informations "11 out of 14,134 vaccinated people got COVID-19 with symptoms. 185 out of 14,073 people who received dummy injections got COVID-19 with symptoms"

# I save the relevant data for the 
vaccine = 14134
placebo = 14073
vaccine_pos = 11
placebo_pos = 185


total = c(rep("V", vaccine), rep("P", placebo)) # V = vaccined
positives = c(rep("P", vaccine_pos), rep("N", vaccine - vaccine_pos), rep("P", placebo_pos), rep("N", placebo - placebo_pos)) # 1 = positive to COVID

# Then I can create the samples list containing these data
data = list(
    total = as.integer(factor(total)),
    length_total = length(unique(total)),
    positives = ifelse(positives == "N", 0, 1),
    length_positives = length(positives)
)

# And then I can use the .bug file to create the posterior
jags_model = jags.model('ex02_les06.bug', data)
update(jags_model, 1000)

# Creation of the rjags chain
moderna_chain = coda.samples(jags_model, c("distribution"), n.iter=10000)
print(summary(moderna_chain))

# And in the end I create a plot for clarity of the results
options(repr.plot.width=20, repr.plot.height=8)
plot(moderna_chain, col="navy")

Compiling model graph
   Resolving undeclared variables
   Allocating nodes
Graph information:
   Observed stochastic nodes: 28207
   Unobserved stochastic nodes: 2
   Total graph size: 56420

Initializing model


Iterations = 1001:11000
Thinning interval = 1 
Number of chains = 1 
Sample size per chain = 10000 

1. Empirical mean and standard deviation for each variable,
   plus standard error of the mean:

                     Mean        SD  Naive SE Time-series SE
distribution[1] 0.0132847 0.0009512 9.512e-06      9.512e-06
distribution[2] 0.0009775 0.0002627 2.627e-06      2.722e-06

2. Quantiles for each variable:

                     2.5%       25%       50%      75%    97.5%
distribution[1] 0.0114651 0.0126417 0.0132704 0.013909 0.015210
distribution[2] 0.0005299 0.0007888 0.0009553 0.001143 0.001556

# For the 95% credibility interval and the efficacy I load both the chain and the result file
chain_df = as.data.frame( as.mcmc(moderna_chain) )
results = summary(moderna_chain)

# 95% CI:
cat("The Credibility interval is [", round(results$quantiles[1,1], 4), ",", round(results$quantiles[1,5], 4),"] for the Vaccined distribution\n")
cat("The Credibility interval is [", round(results$quantiles[2,1], 5), ",", round(results$quantiles[2,5], 5),"] for the Placeboed distribution\n")

# For plotting reasons I retrieve mean, variance and standard deviation of the chain
chain_means = sapply(chain_df, mean)
chain_vars = sapply(chain_df, var)
chain_stds = sqrt(chain_vars)

# Efficacy
cat("The efficacy of the test is ", round((1 - chain_means[2]/chain_means[1])*100, 2), "%")

The Credibility interval is [ 0.0115 , 0.0152 ] for the Vaccined distribution
The Credibility interval is [ 0.00053 , 0.00156 ] for the Placeboed distribution
The Credibility interval is [ 0.00053 , 0.00156 ] for the Placeboed distribution
The efficacy of the test is  92.64 %

These results are in accordance with the one found by EMA, that I quote: "This means that the vaccine demonstrated a 94.1% efficacy in the trial. The trial also showed 90.9% efficacy in participants at risk of severe COVID-19, including those with chronic lung disease, heart disease, obesity, liver disease, diabetes or HIV infection."

I recall that AstraZaneca results are no longer available on the website and that Janssen and Jcovden do have the same database in this moment

Exercise 3

According to the official COVID-19 vaccination data, 70% of the world population has received at least one dose of a COVID-19 vaccine. A global vaccination dataset is available [https://ourworldindata.org/covid-vaccinations]

The European Centre for Disease Prevention and Control published a downloadable file [https://github.com/owid/covid-19-data/tree/master/public/data] containing information on COVID-19 vaccination in the EU/EEA.

Analyze the data and produce the following plots:

• number of vaccinated people (cumulative, daily and week average)
• number of confirmed deaths by COVID-19, both cumulative and weekly average

$\bold{a)}$ The dataset is the same, thus I load it in R as a dataframe and then I can count the number of vaccinated people:

# Load the csv dataset for global vaccination:
filein = "data/owid-covid-data.csv"
data = read.csv(filein, header = T)

# Print the tail of the df
tail(data, 10)

A data.frame: 10 × 67

	iso_code	continent	location	date	total_cases	new_cases	new_cases_smoothed	total_deaths	new_deaths	new_deaths_smoothed	⋯	male_smokers	handwashing_facilities	hospital_beds_per_thousand	life_expectancy	human_development_index	population	excess_mortality_cumulative_absolute	excess_mortality_cumulative	excess_mortality	excess_mortality_cumulative_per_million
	<chr>	<chr>	<chr>	<chr>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	⋯	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>
313257	ZWE	Africa	Zimbabwe	2023-05-15	264848	7	10.286	5690	0	0.286	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313258	ZWE	Africa	Zimbabwe	2023-05-16	264848	0	9.429	5690	0	0.286	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313259	ZWE	Africa	Zimbabwe	2023-05-17	264848	0	8.143	5690	0	0.143	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313260	ZWE	Africa	Zimbabwe	2023-05-18	264848	0	7.000	5690	0	0.143	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313261	ZWE	Africa	Zimbabwe	2023-05-19	264848	0	4.571	5690	0	0.143	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313262	ZWE	Africa	Zimbabwe	2023-05-20	264848	0	3.857	5690	0	0.000	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313263	ZWE	Africa	Zimbabwe	2023-05-21	264848	0	1.000	5690	0	0.000	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313264	ZWE	Africa	Zimbabwe	2023-05-22	264848	0	0.000	5690	0	0.000	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313265	ZWE	Africa	Zimbabwe	2023-05-23	264848	0	0.000	5690	0	0.000	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA
313266	ZWE	Africa	Zimbabwe	2023-05-24	264848	0	0.000	5690	0	0.000	⋯	30.7	36.791	1.7	61.49	0.571	16320539	NA	NA	NA	NA

Looking at the loaded data, I see that there is a location for the world, therefore I use that: ohterwise I could have grouped by the date and summed over all the other columns

# I then select only the required data (using both "iso_code" or "location" as key) # world_data = data[data$iso_code == "OWID_WRL", ]
world_data = data[data$location == "World", ]

library(lubridate)

# Then I get rid of the NA data:
world_data = world_data[!is.na(world_data$total_cases), ]

# And then I convert the date into a real date format and add a column in order to get the number of the week and the year (for further analysis)
world_data$date = as.Date.character((world_data$date), "%Y-%m-%d")
world_data$week = week(world_data$date)
world_data$year = as.numeric(format(world_data$date, "%Y"))

tail(world_data, 10)

Attaching package: ‘lubridate’


The following objects are masked from ‘package:base’:

    date, intersect, setdiff, union

A data.frame: 10 × 69

	iso_code	continent	location	date	total_cases	new_cases	new_cases_smoothed	total_deaths	new_deaths	new_deaths_smoothed	⋯	hospital_beds_per_thousand	life_expectancy	human_development_index	population	excess_mortality_cumulative_absolute	excess_mortality_cumulative	excess_mortality	excess_mortality_cumulative_per_million	week	year
	<chr>	<chr>	<chr>	<date>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	⋯	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>	<dbl>
309538	OWID_WRL		World	2023-05-15	766543531	123350	77506.00	6933813	263	600.143	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	20	2023
309539	OWID_WRL		World	2023-05-16	766583982	40451	75205.14	6934075	262	601.714	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	20	2023
309540	OWID_WRL		World	2023-05-17	766628655	44673	75054.00	6934220	145	588.857	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	20	2023
309541	OWID_WRL		World	2023-05-18	766667443	38788	74006.43	6934391	171	584.143	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	20	2023
309542	OWID_WRL		World	2023-05-19	766724863	57420	75716.86	6934779	388	609.286	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	20	2023
309543	OWID_WRL		World	2023-05-20	766758430	33567	73868.71	6934951	172	604.714	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	20	2023
309544	OWID_WRL		World	2023-05-21	766844213	85783	60576.00	6935527	576	282.429	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	21	2023
309545	OWID_WRL		World	2023-05-22	766889982	45769	49493.00	6935837	310	289.143	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	21	2023
309546	OWID_WRL		World	2023-05-23	766894311	4329	44332.71	6935876	39	257.286	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	21	2023
309547	OWID_WRL		World	2023-05-24	766894311	0	37950.86	6935876	0	236.571	⋯	2.705	72.58	0.737	7975105024	NA	NA	NA	NA	21	2023

Then I can divide the three required cathegories

# I take the cumulative number of vaccined people after removing NAs
if (length(world_data[is.na(world_data$people_vaccinated), ]$people_vaccinated) > 0) {
    world_data[is.na(world_data$people_vaccinated), ]$people_vaccinated = 0}
cumulative_vaccines = world_data$people_vaccinated

# Then the daily vaccined people are similarly retrieved
if (length(world_data[is.na(world_data$new_vaccinations), ]$new_vaccinations) > 0) {
    world_data[is.na(world_data$new_vaccinations), ]$new_vaccinations = 0}
daily_vaccines = world_data$new_vaccinations

# For the weekly average I have to add a column
weekly_vaccines = world_data %>% group_by(week, year) %>% summarise(total_per_week = mean(new_vaccinations))
weekly_vaccines$cumulative_week = as.Date(paste(weekly_vaccines$week, weekly_vaccines$year, 'Sun'), '%U %Y %a')
weekly_vaccines = weekly_vaccines[order(weekly_vaccines$cumulative_week),]

`summarise()` has grouped output by 'week'. You can override using the
`.groups` argument.
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 368 in year 2020 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 366 in year 2021 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 365 in year 2022 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 368 in year 2020 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 366 in year 2021 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 365 in year 2022 is invalid”

# And then I plot the behaviour of these two selections
par(mfrow=c(1,2), oma = c(0, 0, 0, 0))
options(repr.plot.width=20, repr.plot.height=10)

plot(world_data$date, cumulative_vaccines, type='l', lwd=3, col="navy", xlab="date", ylab="# cases", cex=1.5,
    main = paste("Cumulative distribution of the number of vaccined people"))
plot(world_data$date, daily_vaccines, type='l', lwd=3, col="navy", xlab="date", ylab="# cases", cex=1.5, main = paste("New vaccines per day"), ylim = c(0, 6e7))
lines(weekly_vaccines$cumulative_week, weekly_vaccines$total_per_week, type='o', lwd=3, col="red")
legend("topright", legend = c("Daily data", "Weekly averages"), col = c("navy", "red"), lty = c(1, 1), lwd = c(3, 3))

$\bold{b)}$ with a similar code I can count the number of deaths:

# I start again from the "world_data" dataframe

# I take the cumulative number of dead people after removing NAs
if (length(world_data[is.na(world_data$total_deaths), ]$total_deaths) > 0) {
    world_data[is.na(world_data$total_deaths), ]$total_deaths = 0}
cumulative_deaths = world_data$total_deaths

# Then the daily vaccined people are similarly retrieved
if (length(world_data[is.na(world_data$new_deaths), ]$new_deaths) > 0) {
    world_data[is.na(world_data$new_deaths), ]$new_deaths = 0}
daily_deaths = world_data$new_deaths

# For the weekly average I have to add a column
weekly_deaths = world_data %>% group_by(week, year) %>% summarise(total_per_week = mean(new_deaths))
weekly_deaths$cumulative_week = as.Date(paste(weekly_deaths$week, weekly_deaths$year, 'Sun'), '%U %Y %a')
weekly_deaths = weekly_deaths[order(weekly_deaths$cumulative_week),]

`summarise()` has grouped output by 'week'. You can override using the
`.groups` argument.
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 368 in year 2020 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 366 in year 2021 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 365 in year 2022 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 368 in year 2020 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 366 in year 2021 is invalid”
Warning message in strptime(x, format, tz = "GMT"):
“(0-based) yday 365 in year 2022 is invalid”

# And then I plot the behaviour of these two selections
par(mfrow=c(1,2), oma = c(0, 0, 0, 0))
options(repr.plot.width=20, repr.plot.height=10)

plot(world_data$date, cumulative_deaths, type='l', lwd=3, col="navy", xlab="date", ylab="# cases", cex=1.5,
    main = paste("Cumulative distribution of the number of dead people"))
plot(world_data$date, daily_deaths, type='l', lwd=3, col="navy", xlab="date", ylab="# cases", cex=1.5, main = paste("New deaths per day"), ylim = c(0, 2.3e4))
lines(weekly_deaths$cumulative_week, weekly_deaths$total_per_week, type='o', lwd=3, col="red")
legend("topright", legend = c("Daily data", "Weekly averages"), col = c("navy", "red"), lty = c(1, 1), lwd = c(3, 3))